Question

If the radius of a right circular cone is reduce by 10% and its height is increased by 40% , what will be the percentage increase or decrease in its volume?​

Answer 1

Given : The radius of a right circular cone is reduce by 10% and its height is increased by 40% .

Exigency To Find : The percentage increase or decrease in its volume .

⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀

⠀⠀⠀⠀⠀Given that ,

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀The radius of a right circular cone is reduce by 10% .

Therefore,

⠀⠀⠀The new radius ( r ) will be : r - 10 /100r

$\qquad:\implies \sf Radius \:(r)\: =\: r - \dfrac{10}{100}r$

$\qquad:\implies \sf Radius \:(r)\: =\: r - \cancel {\dfrac{10}{100}}r$

$\qquad:\implies \sf Radius \:(r)\: =\: r - 0.1r$

$\qquad:\implies \sf Radius \:(r)\: =\: 0.9r$

$\qquad \therefore \pmb{\underline{\purple{\:Radius \:(r)\: =\: 0.9r }} }\:\:\bigstar$

⠀⠀⠀⠀⠀⠀&

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀The height of Right Circular cone is increased by 40% .

Therefore,

⠀⠀⠀The new height ( h ) will be : h + 40 / 100h

$\qquad:\implies \sf Height \:(h)\: =\: h + \dfrac{40}{100}h$

$\qquad:\implies \sf Height \:(h)\: =\: h + \cancel {\dfrac{40}{100}}h$

$\qquad:\implies \sf Height \:(h)\: =\: h + 0.4 h$

$\qquad:\implies \sf Height \:(h)\: =\: 1.4 h$

$\qquad \therefore \pmb{\underline{\purple{\:Height \:(h)\: =\: 1.4h }} }\:\:\bigstar$

Now ,

$\dag\:\:\frak{ As,\:We\:know\:that\::}\:\\\\ \qquad \maltese \:\bf Volume \:of\:cone \::$

$\qquad \dag\:\:\bigg\lgroup \sf{Volume _{(Cone)} \:: \dfrac{1}{3} \pi r^2 h }\bigg\rgroup$

$\qquad \dashrightarrow \sf Volume _{(Cone)} \:=\: \dfrac{1}{3} \pi r^2 h$

⠀⠀⠀⠀⠀⠀$\underline {\boldsymbol{\star\:Now \: By \: Substituting \: the \: known \: Values \::}}$

$\qquad \dashrightarrow \sf Volume _{(Cone)} \:=\: \dfrac{1}{3} \pi (0.9r)^2 \times (1.4 h )$

$\qquad \dashrightarrow \sf Volume _{(Cone)} \:=\: \dfrac{1}{3} \pi (0.81r^2) \times (1.4 h )$

$\qquad \dashrightarrow \sf Volume _{(Cone)} \:=\: \dfrac{1}{3} \pi(1.134) r^2h$

⠀⠀⠀⠀⠀⠀⠀⠀▪︎ ⠀Now , By Assuming the Volume of Cone ( V ) we get ,

$\qquad \dashrightarrow \sf Volume _{(Cone)} \:=\: 1.134V$

⠀⠀⠀As , The new Volume is in Positive therefore, The Volume should be increased .

Therefore,

⠀⠀⠀⠀⠀⠀Increase in volume ( V ) will be :

$\qquad \dashrightarrow \sf Increase_{(Volume)} \:=\: V - \: 1.134V$

$\qquad \dashrightarrow \sf Increase_{(Volume)} \:=\: 0.134V$

And , Now Increase in Percent will be :

$\qquad \dashrightarrow \sf Increase\:Percentage\:_{(Volume)} \:=\:\dfrac{ 0.134V}{V} \times 100$

$\qquad \dashrightarrow \sf Increase\:Percentage\:_{(Volume)} \:=\:\dfrac{ 0.134\cancel{V}}{\cancel {V}} \times 100$

$\qquad \dashrightarrow \sf Increase\:Percentage\:_{(Volume)} \:=\: 0.134 \times 100$

$\qquad \dashrightarrow \sf Increase\:Percentage\:_{(Volume)} \:=\: 13.4\:\% \:$

$\qquad \therefore \pmb{\underline{\purple{\:Increase\:Percentage\:_{(Volume)} \:=\: 13.4\:\% \: \: }} }\:\:\bigstar$

⠀⠀$\therefore {\underline{ \sf \:Hence, \:Increase \:of\:Percentage \:in \:Volume\:is\:\bf 13.4 \:\% \:\: }}$

Answer 2

Given :

• Shape taken = Cone
• The radius of the cone is reduced by 10% and height is increased by 40%.

Aim :

• To find the increase or decrease in its volume when the radius and height is changed.

Answer :

Formula to use :

$\boxed{\sf \longrightarrow Volume\:of\:a\:cone = \dfrac{1}{3} \times \pi \times (radius)^{2} \times height}$

Let,

• Radius = r units
• Height = h units.

New radius :

If radius is reduced by 10%,

$\implies \sf Decrease\:in\:radius = \dfrac{10}{100} \times r$

$\implies \sf \dfrac{1\not0}{10\not0}$

$\implies \sf \dfrac{r}{10} = Decrease\:in\:radius$

• New radius = Initial (radius) - (Decrease)

$\implies \sf r - \dfrac{r}{10} = \dfrac{9r}{10}$

New height :

if height is increased by 40%,

$\implies \sf Increase\:in\: height = \dfrac{40}{100} \times h$

$\implies \sf \dfrac{4\not0}{10\not0}$

$\implies \sf \dfrac{2h}{5} = Increase\:in\: height$

• New height = (Initial height) + (Increase in height)

$\implies \sf h + \dfrac{2h}{5} = \dfrac{7h}{5}$

Initial volume :

• Radius = r units
• Height = h units

Volume :

$\implies \sf Volume = \dfrac{\pi r^{2}h}{3}$

New volume :

• Radius = 9r/10 units
• Height = 7h/5 units

Volume :

$\implies \sf Volume = \pi \times \dfrac{9r}{10} \times \dfrac{9r}{10} \times \dfrac{7h}{5} \times \dfrac{1}{3}$

$\implies \sf Volume = \dfrac{567 \pi{r}^{2}h }{1500}$

Difference/Increase :

(New volume) - (Initial volume)

$\implies \sf \dfrac{567\pi {r}^{2}h }{1500} - \dfrac{\pi {r}^{2}h }{3}$

LCM = 1500

$\implies \sf \dfrac{567\pi {r}^{2}h - 500\pi {r}^{2}h }{1500}$

$\implies \sf \dfrac{67\pi {r}^{2}h }{1500}$

Increase % :

$\implies \sf \dfrac{Increase}{Initial} \times 100$

$\implies \sf \dfrac{ \dfrac{67\pi {r}^{2}h }{1500} }{ \dfrac{\pi {r}^{2}h }{3} } \times 100$

$\implies \sf \dfrac{67\pi {r}^{2}h }{1500} \div \dfrac{\pi {r}^{2}h }{3} \times 100$

Cancelling πr²h and reducing 3 and 1500 to it's lowest terms,

$\implies \sf \dfrac{67}{500} \times 100$

$\implies \sf \dfrac{67}{5 \not0 \not0} \times 1 \not0 \not0$

$\implies \sf \dfrac{67}{5} = 13.4\%$

Hence, the increase% is 13.4%.